Calculus

Integration of Functions

Integration of Functions Logo

Trigonometric Integrals

  • In this topic, we will study how to integrate certain combinations involving products and powers of trigonometric functions.

    We consider \(8\) cases.

    1. Integrals of the form \({\large\int\normalsize} {\cos ax\cos bxdx} ,\) \({\large\int\normalsize} {\sin ax\cos bxdx} ,\) \({\large\int\normalsize} {\sin ax\sin bxdx}\)

    To evaluate integrals of products of sine and cosine with different arguments, we apply the identities

    • \({\cos ax\cos bx }\) \(= {{\large\frac{1}{2}\normalsize} \left[ {\cos \left( {ax + bx} \right) }\right.} \) \(+\, {\left.{ \cos \left( {ax – bx} \right)} \right]} \)
    • \({\sin ax\cos bx } \) \(= {{\large\frac{1}{2}\normalsize} \left[ {\sin \left( {ax + bx} \right) }\right.} \) \(+\, {\left.{ \sin \left( {ax – bx} \right)} \right]} \)
    • \({\sin ax\sin bx } \) \(= {-{\large\frac{1}{2}\normalsize} \left[ {\cos \left( {ax + bx} \right) }\right.} \) \(-\, {\left.{ \cos \left( {ax – bx} \right)} \right]} \)

    2. Integrals of the form \({\large\int\normalsize} {{\sin^m}x\,{\cos^n}xdx}\)

    We assume here that the powers \(m\) and \(n\) are non-negative integers.

    To find an integral of this form, use the following substitutions:

    1. If \(m\) (the power of sine) is odd, we use the \(u-\)substitution \[{u = \cos x,\;\;}\kern0pt{du = – \sin xdx}\] and the identity \({\sin ^2}x + {\cos ^2}x = 1\) to express the remaining even power of sine in \(u-\)terms.
    2. If \(n\) (the power of cosine) is odd, we use the \(u-\)substitution \[{u = \sin x,\;\;}\kern0pt{du = \cos xdx}\] and the identity \({\sin ^2}x + {\cos ^2}x = 1\) to express the remaining even power of cosine in \(u-\)terms.
    3. If both powers \(m\) and \(n\) are even, we reduce the powers using the the double angle formulas \[{{\sin ^2}x = \frac{{1 – \cos 2x}}{2},\;\;}\kern0pt{{\cos ^2}x = \frac{{1 + \cos 2x}}{2}.}\]

    The integrals of type \(\int {{{\sin }^n}xdx} \) and \(\int {{{\cos }^n}xdx} \) can be evaluated by reduction formulas

    \[{\int {{{\sin }^n}xdx} }={ – \frac{{{{\sin }^{n – 1}}x\cos x}}{n} }+{ \frac{{n – 1}}{n}\int {{{\sin }^{n – 2}}xdx} ,}\]

    \[{\int {{{\cos }^n}xdx} }={ \frac{{{{\cos }^{n – 1}}x\sin x}}{n} }+{ \frac{{n – 1}}{n}\int {{{\cos }^{n – 2}}xdx} .}\]

    3. Integrals of the form \({\large\int\normalsize} {{\tan^n}xdx} \)

    The power of the integrand can be reduced using the trigonometric identity \(1 + {\tan ^2}x = {\sec ^2}x\) and the reduction formula

    \[
    {\int {{\tan^n}xdx} }
    = {\int {{{\tan }^{n – 2}}x\,{{\tan }^2}xdx} }
    = {\int {{{\tan }^{n – 2}}x\left( {{{\sec }^2}x – 1} \right)dx} }
    = {\frac{{{{\tan }^{n – 1}}x}}{{n – 1}} }-{ \int {{{\tan }^{n – 2}}xdx} .}
    \]

    4. Integrals of the form \({\large\int\normalsize} {{{\cot }^n}xdx} \)

    The power of the integrand can be reduced using the trigonometric identity \(1 + {\cot ^n}x = {\csc ^n}x\) and the reduction formula

    \[
    {\int {{{\cot }^n}xdx} }
    = {\int {{{\cot }^{n – 2}}x\,{{\cot }^2}xdx} }
    = {\int {{{\cot }^{n – 2}}x\left( {{{\csc }^2}x – 1} \right)dx} }
    = { – \frac{{{{\cot }^{n – 1}}x}}{{n – 1}} }-{ \int {{{\cot }^{n – 2}}xdx} .}
    \]

    5. Integrals of the form \({\large\int\normalsize} {{\sec^n}xdx} \)

    This type of integrals can be simplified with help of the reduction formula:

    \[
    {\int {{\sec^n}xdx} }
    = {\frac{{{{\sec }^{n – 2}}x\tan x}}{{n – 1}} }+{ \frac{{n – 2}}{{n – 1}}\int {{\sec^{n – 2}}xdx} .}
    \]

    6. Integrals of the form \({\large\int\normalsize} {{\csc^n}xdx} \)

    Similarly to the previous examples, this type of integrals can be simplified by the formula

    \[
    {\int {{\csc^n}xdx} }
    = { – \frac{{{\csc^{n – 2}}x \cot x}}{{n – 1}} }+{ \frac{{n – 2}}{{n – 1}}\int {{\csc^{n – 2}}xdx} .}
    \]

    7. Integrals of the form \({\large\int\normalsize} {{\tan^m}x\,{\sec^n}xdx} \)

    1. If the power of the secant \(n\) is even, then using the identity \(1 + {\tan ^2}x \) \(= {\sec ^2}x\) the secant function is expressed as the tangent function. The factor \({\sec ^2}x\) is separated and used for transformation of the differential. As a result, the entire integral (including differential) is expressed in terms of the function \(\tan x.\)
    2. If both the powers \(n\) and \(m\) are odd, then the factor \(\sec x \tan x,\) which is necessary to transform the differential, is separated. Then the entire integral is expressed in terms of \(\sec x.\)
    3. If the power of the secant \(n\) is odd, and the power of the tangent \(m\) is even, then the tangent is expressed in terms of the secant using the identity \(1 + {\tan ^2}x \) \(= {\sec ^2}x.\) After this substitution, you can calculate the integrals of the secant.

    8. Integrals of the form \({\large\int\normalsize} {{\cot^m}x\,{\csc^n}xdx} \)

    1. If the power of the cosecant \(n\) is even, then using the identity \(1 + {\cot^2}x = {\csc ^2}x\) the cosecant function is expressed as the cotangent function. The factor \({\csc^2}x\) is separated and used for transformation of the differential. As a result, the integrand and differential are expressed in terms of \(\cot x.\)
    2. If both the powers \(n\) and \(m\) are odd, then the factor \(\cot x \csc x,\) which is necessary to transform the differential, is separated. Then the integral is expressed in terms of \(\csc x.\)
    3. If the power of the cosecant \(n\) is odd, and the power of the cotangent \(m\) is even, then the cotangent is expressed in terms of the cosecant using the identity \(1 + {\cot^2}x = {\csc ^2}x.\) After this substitution, you can find the integrals of the cosecant.

  • Solved Problems

    Click a problem to see the solution.

    Example 1

    Calculate the integral \({\large\int\normalsize} {{\sin^3}xdx}.\)

    Example 2

    Evaluate the integral \({\large\int\normalsize} {{\cos^5}xdx}.\)

    Example 3

    Find the integral \({\large\int\normalsize} {{\sin^6}xdx}.\)

    Example 4

    Find the integral \(\int {{{\sin }^2}x\,{{{\cos }^3}x}dx}.\)

    Example 5

    Calculate the integral \({\large\int\normalsize} {{{\sin }^2}x\,{{\cos }^4}xdx}.\)

    Example 6

    Evaluate the integral \(\int {{{\sin }^3}x\,{{\cos }^4}xdx}.\)

    Example 7

    Evaluate the integral \(\int {{{\sin }^3}x\,{{\cos }^5}xdx}.\)

    Example 8

    Evaluate the integral \(\int {{{\sin }^3}x\,{{\cos }^3}xdx}.\)

    Example 9

    Find the integral \({\large\int\normalsize} {{{\sin }^3}x\sqrt {\cos x} dx}.\)

    Example 10

    Evaluate the integral \(\int {\sin 2x\cos 5xdx}.\)

    Example 11

    Evaluate the integral \({\large\int\normalsize} {\sin {\large\frac{x}{4}\normalsize} \cos {\large\frac{x}{3}\normalsize} dx}.\)

    Example 12

    Evaluate the integral \(\int {\cos \large{\frac{x}{2}}\normalsize\cos \large{\frac{x}{3}}\normalsize dx}.\)

    Example 13

    Evaluate the integral \(\int {\sin \large{\frac{x}{3}}\normalsize\sin \large{\frac{x}{4}}\normalsize dx}.\)

    Example 14

    Evaluate the integral \({\large\int\normalsize} {{{\tan }^4}xdx}.\)

    Example 15

    Find the integral \(\int {{{\cot }^4}xdx}.\)

    Example 16

    Find the integral \(\int {{{\tan }^5}xdx}.\)

    Example 17

    Calculate the integral \({\large\int\normalsize} {{{\cot }^5}xdx}.\)

    Example 18

    Calculate the integral \({\large\int\normalsize} {{{\sec }^3}xdx}.\)

    Example 19

    Find the integral \(\int {{{\csc }^3}xdx}.\)

    Example 20

    Evaluate the integral \({\large\int\normalsize} {{{\csc }^4}xdx}.\)

    Example 21

    Find the integral \(\int {{{\sec }^4}xdx}.\)

    Example 22

    Compute \({\large\int\normalsize} {{\tan^3}x\,{{\sec }^2}xdx}.\)

    Example 23

    Compute \({\large\int\normalsize} {{\tan^2}x\sec xdx}.\)

    Example 1.

    Calculate the integral \({\large\int\normalsize} {{\sin^3}xdx}.\)

    Solution.

    Let \(u = \cos x,\) \(du = -\sin xdx.\) Then

    \[
    {\int {{\sin^3}xdx} }={ \int {{\sin^2}x\sin xdx} }
    = {\int {\left( {1 – {\cos^2}x} \right)\sin xdx} }
    = { – \int {\left( {1 – {u^2}} \right)du} }
    = {\int {\left( {{u^2} – 1} \right)du} }
    = {\frac{{{u^3}}}{3} – u + C }
    = {{\frac{{{{\cos }^3}x}}{3} – \cos x }+{ C.}}
    \]

    Example 2.

    Evaluate the integral \({\large\int\normalsize} {{\cos^5}xdx}.\)

    Solution.

    Making the substitution \(u = \sin x,\) \(du = \cos xdx\) and using the identity \({\cos ^2}x = 1 – {\sin ^2}x,\) we obtain

    \[
    {\int {{\cos^5}xdx} }
    = {\int {{{\left( {{\cos^2}x} \right)}^2}\cos xdx} }
    = {\int {{{\left( {1 – {{\sin }^2}x} \right)}^2}\cos x dx} }
    = {\int {{{\left( {1 – {u^2}} \right)}^2}du} }
    = {\int {\left( {1 – 2{u^2} + {u^4}} \right)du} }
    = {{u – \frac{{2{u^3}}}{3} + \frac{{{u^5}}}{5} }+{ C }}
    = {{\sin x – \frac{{2{{\sin }^3}x}}{3} }+{ \frac{{{{\sin }^5}x}}{5} }+{ C.}}
    \]

    Example 3.

    Find the integral \({\large\int\normalsize} {{\sin^6}xdx}.\)

    Solution.

    Using identities \({\sin ^2}x = {\large\frac{{1 – \cos 2x}}{2}\normalsize}\) and \({\cos ^2}x = {\large\frac{{1 + \cos 2x}}{2}\normalsize},\) we can write:

    \[ {I = \int {{\sin^6}xdx} } = {\int {{{\left( {{\sin^2}x} \right)}^3}dx} } = {\frac{1}{8}\int {{{\left( {1 – \cos 2x} \right)}^3}dx} } = {{\frac{1}{8}\int {\left( {1 – 3\cos 2x }\right.}}}+{{{\left.{ 3{{\cos }^2}2x – {{\cos }^3}2x} \right)dx} }} = {\frac{x}{8} – \frac{3}{8} \cdot \frac{{\sin 2x}}{2} } + {\frac{3}{8}\int {{\cos^2}2xdx} } – {\frac{3}{8}\int {{\cos^3}2xdx}.} \]

    Calculate the integrals in the latter expression.

    \[
    {\int {{\cos^2}2xdx} }
    = {\int {\frac{{1 + \cos 4x}}{2}dx} }
    = {\frac{1}{2}\int {\left( {1 + \cos 4x} \right)dx} }
    = {\frac{1}{2}\left( {x + \frac{{\sin 4x}}{4}} \right) }
    = {\frac{x}{2} + \frac{{\sin 4x}}{8}.}
    \]

    To find the integral \({\large\int\normalsize} {{\cos^3}2xdx},\) we make the substitution \(u = \sin 2x,\) \(du =\) \( 2\cos 2xdx.\) Then

    \[
    {\int {{\cos^3}2xdx} }
    = {\frac{1}{2}\int {2{{\cos }^2}2x\cos 2xdx} }
    = {\frac{1}{2}\int {2\left( {1 – {{\sin }^2}2x} \right)\cos 2xdx} }
    = {\frac{1}{2}\int {\left( {1 – {u^2}} \right)du} }
    = {\frac{u}{2} – \frac{{{u^3}}}{6} }
    = {\frac{{\sin 2x}}{2} – \frac{{{{\sin }^3}2x}}{6}.}
    \]

    Hence, the initial integral is

    \[ {I = \frac{x}{8} – \frac{{3\sin 2x}}{{16}} } + {\frac{3}{8}\left( {\frac{x}{2} + \frac{{\sin 4x}}{8}} \right) } – {\frac{1}{8}\left( {\frac{{\sin 2x}}{2} – \frac{{{{\sin }^3}2x}}{6}} \right) }+{ C } = {\frac{{5x}}{{16}} – \frac{{\sin 2x}}{4} } + {\frac{{3\sin 4x}}{{64}} } + {\frac{{{{\sin }^3}2x}}{{48}} }+{ C.} \]

    Example 4.

    Find the integral \(\int {{{\sin }^2}x\,{{{\cos }^3}x}dx}.\)

    Solution.

    The power of cosine is odd, so we make the substitution

    \[{u = \sin x,\;\;}\kern0pt{du = \cos xdx.}\]

    We rewrite the integral in terms of \(\sin x\) to obtain:

    \[{\int {{{\sin }^2}x\,{{\cos }^3}xdx} }={ \int {{{\sin }^2}x\,{{{\cos }^2}x}\cos xdx} }={ \int {{{\sin }^2}x{\left( {1 – {{\sin }^2}x} \right)}\cos xdx} }={ \int {{u^2}\left( {1 – {u^2}} \right)du} }={ \int {\left( {{u^2} – {u^4}} \right)du} }={ \frac{{{u^3}}}{3} – \frac{{{u^5}}}{5} + C }={ \frac{{{{\sin }^3}x}}{3} – \frac{{{{\sin }^5}x}}{5} + C.}\]

    Example 5.

    Calculate the integral \({\large\int\normalsize} {{{\sin }^2}x\,{{\cos }^4}xdx}.\)

    Solution.

    We can write:

    \[
    {I = \int {{{\sin }^2}x\,{{\cos }^4}xdx} }
    = {\int {{{\left( {\sin x\cos x} \right)}^2}{{\cos }^2}xdx} .}
    \]

    We convert the integrand using the identities

    \[
    {\sin x\cos x = \frac{{\sin 2x}}{2},\;\;\;}\kern-0.3pt
    {{\cos ^2}x = \frac{{1 + \cos 2x}}{2},\;\;\;}\kern-0.3pt
    {{\sin ^2}x = \frac{{1 – \cos 2x}}{2}.}
    \]

    This yields

    \[
    {I = \int {{{\left( {\frac{{\sin 2x}}{2}} \right)}^2}\frac{{1 + \cos 2x}}{2}dx} }
    = {\frac{1}{8}\int {{{\sin }^2}2x\left( {1 + \cos 2x} \right)dx} }
    = {{\frac{1}{8}\int {{{\sin }^2}2xdx} }+{ \frac{1}{8}\int {{{\sin }^2}2x\cos 2xdx} }}
    = {{\frac{1}{8}\int {\frac{{1 – \cos 4x}}{2}dx} }}+{{ \frac{1}{{16}}\int {2{{\sin }^2}2x\cos 2xdx} }}
    = {{\frac{1}{{16}}\int {\left( {1 – \cos 4x} \right)dx} }}+{{ \frac{1}{{16}}\int {{{\sin }^2}2x\,d\left( {\sin 2x} \right)} }}
    = {{\frac{1}{{16}}\left( {x – \frac{{\sin 4x}}{4}} \right) }}+{{ \frac{1}{{16}} \cdot \frac{{{{\sin }^3}2x}}{3} }+{ C }}
    = {{\frac{x}{{16}} – \frac{{\sin 4x}}{{64}} }+{ \frac{{{{\sin }^3}2x}}{{48}} }+{ C.}}
    \]

    Example 6.

    Evaluate the integral \(\int {{{\sin }^3}x\,{{\cos }^4}xdx}.\)

    Solution.

    As the power of sine is odd, we use the substitution

    \[{u = \cos x,\;\;}\kern0pt{du = – \sin xdx.}\]

    The integral is written as

    \[{I = \int {{{\sin }^3}x\,{{\cos }^4}xdx} }={ \int {{{\sin }^2}x\,{{\cos }^4}x\sin xdx} .}\]

    By the Pythagorean identity,

    \[{\sin ^2}x = 1 – {\cos ^2}x.\]

    Hence

    \[{I = \int {{{\sin }^2}x\,{{\cos }^4}x\sin xdx} }={ \int {\left( {1 – {{\cos }^2}x} \right){{\cos }^4}x\sin xdx} }={ – \int {\left( {1 – {u^2}} \right){u^4}du} }={ \int {\left( {{u^6} – {u^4}} \right)du} }={ \frac{{{u^7}}}{7} – \frac{{{u^5}}}{5} + C }={ \frac{{{{\cos }^7}x}}{7} – \frac{{{{\cos }^5}x}}{5} + C.}\]

    Example 7.

    Evaluate the integral \(\int {{{\sin }^3}x\,{{\cos }^5}xdx}.\)

    Solution.

    We see that both powers are odd, so we can substitute either \(u = \sin x\) or \(u = \cos x.\) Choosing the least exponent, we have

    \[{u = \cos x,\;\;}\kern0pt{du = – \sin xdx.}\]

    The integral takes the form

    \[{I = \int {{{\sin }^3}x\,{{\cos }^5}xdx} }={ \int {{{\sin }^2}x\,{{\cos }^5}x\sin xdx} .}\]

    Using the Pythagorean identity,

    \[{\sin ^2}x = 1 – {\cos ^2}x,\]

    we can write

    \[{I = \int {{{\sin }^2}x\,{{\cos }^5}x\sin xdx} }={ \int {\left( {1 – {{\cos }^2}x} \right){{\cos }^5}x\sin xdx} }={ – \int {\left( {1 – {u^2}} \right){u^5}du} }={ \int {\left( {{u^7} – {u^5}} \right)du} }={ \frac{{{u^8}}}{8} – \frac{{{u^6}}}{6} + C }={ \frac{{{{\cos }^8}x}}{8} – \frac{{{{\cos }^6}x}}{6} + C.}\]

    Example 8.

    Evaluate the integral \(\int {{{\sin }^3}x\,{{\cos }^3}xdx}.\)

    Solution.

    The powers of both sine and cosine are odd. Hence we can use the substitution \(u = \sin x\) or \(u = \cos x.\) Let’s apply the substitution \(u = \sin x.\) Then \(du = \cos x dx,\) and the integral becomes

    \[{I = \int {{{\sin }^3}x\,{{\cos }^3}xdx} }={ \int {{{\sin }^3}x\,{{\cos }^2}x\cos xdx} .}\]

    By the Pythagorean identity,

    \[{\sin ^2}x = 1 – {\cos ^2}x,\]

    so we obtain

    \[{I = \int {{{\sin }^3}x\,{{\cos }^2}x\cos xdx} }={ \int {{{\sin }^3}x\left( {1 – {{\sin }^2}x} \right)\cos xdx} }={ \int {{u^3}\left( {1 – {u^2}} \right)du} }={ \int {\left( {{u^3} – {u^5}} \right)du} }={ \frac{{{u^4}}}{4} – \frac{{{u^6}}}{6} + C }={ \frac{{{{\sin }^4}x}}{4} – \frac{{{{\sin }^6}x}}{6} + C.}\]

    Page 1
    Problems 1-8
    Page 2
    Problems 9-23